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DENIUS [597]
4 years ago
7

Chris had a piece of rope that was 15 inches long. He wanted to cut the rope into pieces that were each 3/5 of an inch long. How

many pieces of rope can he cut from the original rope?
Mathematics
2 answers:
kati45 [8]4 years ago
4 0

Hey there! :)

Answer:

25 pieces.

Step-by-step explanation:

To solve for the amount of pieces that can be cut, we must divide fractions:

15 ÷ 3/5 = x

When dividing fractions, you must multiply by the reciprocal. Therefore:

\frac{15}{1}* \frac{5}{3} =   \frac{75}{3}  = 25

Therefore, 25 pieces can be cut.

Rashid [163]4 years ago
3 0

Answer:

25 3/5 inch long pieces

Step-by-step explanation:

If you have 15 inches and you cut it into less than inch pieces, you know for sure that it will be more than 15 but less than 30 because the peices are more than half an inch.

to find the answer, you divide 15 by 3/5 because each piece is 3/5 inches long

when you put this into a calculator 15/(3/5) you will get the answer 25.

So, ANSWER = 25 pieces of rope that are 3/5 inches long

Hope this helps and you will understand if you face a problem similar to this :)

You might be interested in
Find the diameter of a circle whose circumference is 74cm
jenyasd209 [6]
  • <em>Answer:</em>

<em>≈ 24 cm</em>

  • <em>Step-by-step explanation:</em>

<em>Hi there !</em>

<em>c = 2πr  } => c = πd => d = c/π</em>

<em>2r = d</em>

<em>replace c ; π</em>

<em>π  =3.14</em>

<em>d = 74cm/3.14</em>

<em>= 23.5668</em>

<em>≈ 24 cm</em>

<em>Good luck !</em>

4 0
4 years ago
A self at a bookstore displays 27 books. Of these 27 books, 9 of the books are nonfiction books. The store owner adds 6 new fict
mrs_skeptik [129]
Right now, the ratio of nonfiction to fiction is 9:18.  If this is reduced, it is a ratio of 1:2, so there needs to be twice as many fiction books as nonfiction.

If we add 6 fiction books, we have 24 fiction books.  We can now set up this equation:

\frac{1}{2} = \frac{x}{24}

Now we cross multiply

2x=24

We can solve for x by dividing both sides by two, so:

x = 12.  There will be 12 nonfiction books.
8 0
3 years ago
Help please it's number lines​
skelet666 [1.2K]

sin(θ+β)=−

5

7

−4

15

2

Step-by-step explanation:

step 1

Find the sin(\theta)sin(θ)

we know that

Applying the trigonometric identity

sin^2(\theta)+ cos^2(\theta)=1sin

2

(θ)+cos

2

(θ)=1

we have

cos(\theta)=-\frac{\sqrt{2}}{3}cos(θ)=−

3

2

substitute

sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1sin

2

(θ)+(−

3

2

)

2

=1

sin^2(\theta)+ \frac{2}{9}=1sin

2

(θ)+

9

2

=1

sin^2(\theta)=1- \frac{2}{9}sin

2

(θ)=1−

9

2

sin^2(\theta)= \frac{7}{9}sin

2

(θ)=

9

7

sin(\theta)=\pm\frac{\sqrt{7}}{3}sin(θ)=±

3

7

Remember that

π≤θ≤3π/2

so

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

sin(\theta)=-\frac{\sqrt{7}}{3}sin(θ)=−

3

7

step 2

Find the sec(β)

Applying the trigonometric identity

tan^2(\beta)+1= sec^2(\beta)tan

2

(β)+1=sec

2

(β)

we have

tan(\beta)=\frac{4}{3}tan(β)=

3

4

substitute

(\frac{4}{3})^2+1= sec^2(\beta)(

3

4

)

2

+1=sec

2

(β)

\frac{16}{9}+1= sec^2(\beta)

9

16

+1=sec

2

(β)

sec^2(\beta)=\frac{25}{9}sec

2

(β)=

9

25

sec(\beta)=\pm\frac{5}{3}sec(β)=±

3

5

we know

0≤β≤π/2 ----> II Quadrant

so

sec(β), sin(β) and cos(β) are positive

sec(\beta)=\frac{5}{3}sec(β)=

3

5

Remember that

sec(\beta)=\frac{1}{cos(\beta)}sec(β)=

cos(β)

1

therefore

cos(\beta)=\frac{3}{5}cos(β)=

5

3

step 3

Find the sin(β)

we know that

tan(\beta)=\frac{sin(\beta)}{cos(\beta)}tan(β)=

cos(β)

sin(β)

we have

tan(\beta)=\frac{4}{3}tan(β)=

3

4

cos(\beta)=\frac{3}{5}cos(β)=

5

3

substitute

(4/3)=\frac{sin(\beta)}{(3/5)}(4/3)=

(3/5)

sin(β)

therefore

sin(\beta)=\frac{4}{5}sin(β)=

5

4

step 4

Find sin(θ+β)

we know that

sin(A + B) = sin A cos B + cos A sin Bsin(A+B)=sinAcosB+cosAsinB

so

In this problem

sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)sin(θ+β)=sin(θ)cos(β)+cos(θ)sin(β)

we have

sin(\theta)=-\frac{\sqrt{7}}{3}sin(θ)=−

3

7

cos(\theta)=-\frac{\sqrt{2}}{3}cos(θ)=−

3

2

sin(\beta)=\frac{4}{5}sin(β)=

5

4

cos(\beta)=\frac{3}{5}cos(β)=

5

3

substitute the given values in the formula

sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})sin(θ+β)=(−

3

7

)(

5

3

)+(−

3

2

)(

5

4

)

sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})sin(θ+β)=(−3

15

7

)+(−4

15

2

)

sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}sin(θ+β)=−

5

7

−4

15

2

Step-by-step explanation:

i hope it helps to you

5 0
3 years ago
Find the equation of the tangent line to the curve when x has the given value.
Leviafan [203]

Answer:

14) The equation of the tangent line to the curve f(x)=-2-x^2 at x = -1 is y=2x-1

15) The rate of learning at the end of eight hours of instruction is w'(8) = 5 \frac{items}{hour}

Step-by-step explanation:

14) To find the equation of a tangent line to a curve at an indicated point you must:

1. Find the first derivative of f(x)

f(x)=-2-x^2\\\\\frac{d}{dx} f(x)=\frac{d}{dx}(-2-x^2)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx} f(x)=\frac{d}{dx}\left(-2\right)-\frac{d}{dx}\left(x^2\right)\\\\f'(x)=-2x

2. Plug x value of the indicated point, x = -1, into f '(x) to find the slope at x.

f'(-1)=-2(-1)=2

3. Plug x value into f(x) to find the y coordinate of the tangent point

f(-1)=-2-(-1)^2=-3

4. Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line

y-y_1=m(x-x_1)\\y+3=2(x+1)\\y=2x-1

5. Graph your function and the equation of the tangent line to check the results.

15) To find the rate of learning at the end of eight hours of instruction you must:

1. Find the first derivative of f(x)

w(t)=15\sqrt[3]{t^2} \\\\\frac{d}{dt}w= \frac{d}{dt}(15\sqrt[3]{t^2})\\\\w'(t)=15\frac{d}{dt}\left(\sqrt[3]{t^2}\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\f=\sqrt[3]{u},\:\:u=\left(t^2\right)\\\\w'(t)=15\frac{d}{du}\left(\sqrt[3]{u}\right)\frac{d}{dt}\left(t^2\right)\\\\w'(t)=15\cdot \frac{1}{3u^{\frac{2}{3}}}\cdot \:2t\\\\\mathrm{Substitute\:back}\:u=\left(t^2\right)

w'(t)=15\cdot \frac{1}{3(t^2)^{\frac{2}{3}}}\cdot \:2t\\w'(t)=\frac{10t}{\left(t^2\right)^{\frac{2}{3}}}

2. Evaluate the derivative a t = 8

w'(t)=\frac{10t}{\left(t^2\right)^{\frac{2}{3}}}\\\\w'(8)=\frac{10\cdot 8}{\left(8^2\right)^{\frac{2}{3}}}\\\\=\frac{80}{\left(8^2\right)^{\frac{2}{3}}}\\\\\left(8^2\right)^{\frac{2}{3}}=16\\\\=\frac{80}{16}\\\\w'(8) = 5 \frac{items}{hour}

The rate of learning at the end of eight hours of instruction is w'(8) = 5 \frac{items}{hour}

7 0
3 years ago
2) Find x if IW = x -2 and IU = 2x - 6<br> H<br> U<br> A) 6.6<br> C) 7<br> B) 5<br> D) 4
Mrac [35]

Given:

The figure of a triangle.

IW= x-2

IU=2x-6

To find:

The value of x

Solution:

From the given figure it is clear that VI is a median of the given triangle. It means I divides UW is two equal parts.

IU=IW

2x-6=x-2

2x-x=6-2

[tex]x=4/tex]

The value of x is 4. Therefore, the correct option is D.

5 0
3 years ago
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